I am supposed to work out which product is formed in which solvent when naphthalen-1-olate is treated with benzyl iodide.
DMSO is a polar aprotic solvent, as it does not possess an acidic hydrogen. So, the negative charge on oxygen will be concentrated as the anion will be poorly solvated. Hence, it should act as a hard nucleophile. Since alkyl halides such as BnI are soft electrophiles, I think that the substituent should preferentially add to the ring.
In 2,2,2-trifluoroethan-1-ol (TFE), I think the reaction would proceed via an SN1 mechanism. Now, the carbocation is a hard electrophile, and thus O-alkylation might be observed.
But considering the better solvation of the negatively charged oxygen in TFE, it would also be more hindered, which suggests that C-alkylation might also be seen.
Which products are actually formed, and am I on the right track?